The empirical formula is C₃H₅ IO₂
Explanation:
Assume 100 g of the compound is present. This changes the percents to grams:
C = 18 g, H = 2.5 g , I = 63. 5 g , O = 16 g
Convert the masses to moles:
C = 18/ 12 = 1.5
H = 2.5/1 = 2.5
I = 63.5/ 127 =0.5
O = 16/16 = 1
Divide by the lowest, seeking the smallest whole-number ratio:
C = 1.5 / 0.5 = 3
H = 2.5/0.5=5
I = 0.5/0.5= 1
O = 1/0.5= 2
Now we can write the Empirical formula as,
C₃H₅ IO₂
